Problem: $h(x) = \begin{cases} \dfrac{1}{x-5} &, & x = 0 \\\\ -(x-4)^2&, & x = 7\\\\ 3x-2&, & x \neq 0,7\end{cases}$ $h(6)=$
Solution: The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $x={6}$. Finding the appropriate assignment rule Since ${6}\neq0$ and ${6} \neq 7$, we should use the third assignment rule $3x-2$. The answer $h({6})= 3 \cdot {6}-2=16$ In conclusion, $h(6)=16$.